Question: Solve for $x$ and $y$ using elimination. ${3x-y = 16}$ ${5x+y = 32}$
Solution: We can eliminate $y$ by adding the equations together when the $y$ coefficients have opposite signs. Add the equations together. Notice that the terms $-y$ and $y$ cancel out. $8x = 48$ $\dfrac{8x}{{8}} = \dfrac{48}{{8}}$ ${x = 6}$ Now that you know ${x = 6}$ , plug it back into $\thinspace {3x-y = 16}\thinspace$ to find $y$ ${3}{(6)}{ - y = 16}$ $18-y = 16$ $18{-18} - y = 16{-18}$ $-y = -2$ $\dfrac{-y}{{-1}} = \dfrac{-2}{{-1}}$ ${y = 2}$ You can also plug ${x = 6}$ into $\thinspace {5x+y = 32}\thinspace$ and get the same answer for $y$ : ${5}{(6)}{ + y = 32}$ ${y = 2}$